Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{10(r - 8)}{6r} \times \dfrac{-2}{8r - 64} $
When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 10(r - 8) \times -2 } { 6r \times (8r - 64) } $ $ z = \dfrac {-2 \times 10(r - 8)} {6r \times 8(r - 8)} $ $ z = \dfrac{-20(r - 8)}{48r(r - 8)} $ We can cancel the $r - 8$ so long as $r - 8 \neq 0$ Therefore $r \neq 8$ $z = \dfrac{-20 \cancel{(r - 8})}{48r \cancel{(r - 8)}} = -\dfrac{20}{48r} = -\dfrac{5}{12r} $